If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content

# Cross product and torque

## Video transcript

all of the torque problems I've done so far on the physics playlist we only just figured out the magnitude of torque but well frankly because that's what normally matters but torque is actually a vector and it's Direction can be found and that is because torque is defined as the cross-product between the radial distance from your axis of rotation and the rotational force being applied so these are both vectors so let's take a look at how I taught you a vectors the first time and then I'll show you how that's really the same thing as what we're doing here with the cross-product except now with the cross-product besides just the magnitude for torque we're also getting the direction but then we'll also see that that direction is a little bit it's just kind of the definition of the direction of torque I don't know how intuitive it really is but what did I teach you before about torque well I said if I had a let's say I had some arm say that's the arm and let's say it's it's you know this could be the hand of a clock or it's pinned down to the wall there so it would rotate around this object and then let's let me switch colors the magenta is getting a little obnoxious let's say it's some distance R from the pivot let's say that distance is 10 all right this is the same thing as R and the magnitude of R is equal to 10 at some distance 10 from the pivot I apply some force F and F I will do in yellow I apply some force F let me do it let me draw it straight I apply some force F at some angle that's my force F it's also a vector has magnitude and direction and let's say let's say that this is 10 meters and let's say that I apply a force of I don't know 7 Newtons let me let me make it more interesting let's say I apply a force of square root of 3 nu and I just threw that out there because I think the numbers will all work out and let's say that the angle between my force and the lever arm or the arm that's rotating let's stick to radians this time stick to radians let's say it's PI over 3 but if you need to visualize that that's 60 degrees PI over 3 radians because a it's equal to theta and so just just based on what we already know about moments or torque what is the what is the torque around this pivot or how much torque is being applied by this force and when we learn torque I'm going to learn moments we realize really the only hard part about these problems is that you don't just multiply the entire rotational force times the distance from the axis of rotation you have to multiply the component of that force that is actually doing the rotation or the component of the force that is perpendicular to this rotating arm or perpendicular to this moment arm so how do we figure that out well the component of this of this force that is perpendicular to this arm I can visually draw it here see it would be looks something like this like that I could draw it there I could also draw it here right that would be this would be the component or this would be the component that is perpendicular to this rotating arm and the component that is parallel would be this but we don't care about that that's kind of not contributing to the rotation the only thing is contributing the rotation is this component of the force and what is the magnitude of this vector right here the component of vector F that is perpendicular to this arm well if this angle let me draw a little triangle down here if this is square root of 3 this is PI over 3 radians or 60 degrees then this is and this is a right angle it's PI over 3 and it's hard to read what is this length right here what's a 30-60-90 triangle and we know that this length here I mean there's a couple of ways you could think about we know trigonometry we know that this is just a square root of three times the sine of PI over three or the sine of 60 degrees and so that equals the square root of three sine of PI over three or a sine of 60 degrees is square root of three over two so square root of three times the square root of 3 is just 3 so that equals three halves so the magnitude of this force vector that is perpendicular the component that is perpendicular to the arm is three over two Newtons and now we can figure out the magnitude of the torque it's three over two Newton's times ten meters so we know the magnitude of the torque and I'm being a little bit more careful with my notation right now to remind you the torque actually is a vector or you can almost view it as a they use this term pseudo vector because it's kind of a well anyway I won't go into that so what is the magnitude of the torque vector well it's three-halves newtons times the distance and I remember I just drew this vector here just to show you the component I could just shift the vector here because this is actually where the force is being applied you could draw that same vector here because you can shift vectors around so this is also three 1/2 Newtons and maybe that makes a little bit clearer so it's three-halves Newtons and times the distance that you are from your pivot arm so times 10 meters and so that is equal to what 15 Newton meters so the magnitude of the torque is 15 Newton meters but all we did now and hopefully this this looks a little bit familiar this is what we learned when we learned moments in torque but all we did now is we figured out the the magnitude of the torque but what if we wanted to know the direction and that's where the cross-product comes in so what was the definition of the cross-product cross product R cross F that is equal to magnitude of R times the magnitude of F times sign of the smallest angle between them times some vector that is perpendicular to both and this is really where it's going to help because all of these right here these are all scalar quantities right so these will specify the direction this but the direction is completely specified by this unit vector and a unit vector is just a vector of magnitude one that's pointing in some direction well look this cross-product this part of it the part that just gives us magnitudes we just calculated that using what we knew before of torques the magnitude of our force vector times sine of theta that gave us the component of the force vector that is perpendicular to the arm and we just multiply that times the magnitude of R and we got the magnitude of the torque vector which was 1515 we could leave out the Newton the Newton meters for now 15 and then its direction is this vector that we specify by n we could call the normal vector and what do we know about this vector it's perpendicular to both R this is R right this is our and it's perpendicular F and the only way that I can visualize in our three-dimensional universe a vector that's perpendicular to both this and this if it pops in or out of this page right because both of these vectors are in the plane that are defined by our video so the if I'm if I'm a vector that's perpendicular to your screen whatever you're watching this on then it's going to be perpendicular to both of these vectors and how do we figure out if that vector pops out or pops into the page we use the right hand rule right the right hand rule we take R is our index finger F is our middle finger and then whichever direction our thumb points in tells us whether or not we are the direction of the the cross product so let's draw it let me see if I can do a good job right here so if that is my index finger that's my index finger and you can kind of imagine like the your your hand sitting on top of this screen right so that's that's my index finger representing our and this is my right hand remember it only works with your right hand if you do your left hand is going to be the opposite and then my my middle finger is going to go in the direction of F and then the rest of my fingers are and I encourage you to draw this so if I were to draw it let me draw my nails just so you know what this is so this is the nail on my index finger that's the nail on my middle finger and so in this situation where as my thumb I'm going to be my thumb is going to be popping out I wish I could and that's the nail of my thumb hopefully that makes some sense right that's the paw might sound at the other side of my and then I could keep drawing but hopefully that makes some some sense this is my index finger this the middle finger my thumb is pointing out of the page so that tells us that the torque is actually pointing out of the page so that the direction of this unit vector n is going to be out of the page and we could signify that by a circle with the dot and I'm almost at my time limit and so there you have it torque is applied to or the cross product as it is applied to torque see in the next video